Difference between revisions of "022 Exam 2 Sample B, Problem 7"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1:  
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!(a) Step 1:  
 
|-
 
|-
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|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. After substitution we have
+
|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. Substituting, we have
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::<math>\int x e^{3x^2+1}\,dx = \frac{1}{6} \int e^{u}\, du. </math>
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::<math>\int x e^{3x^2+1}\,dx \,=\,  \int xe^{u}\cdot\frac{du}{6}\,=\,\frac{1}{6}\int e^u\,du\,=\,\frac{1}{6}u. </math>
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|-
 
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|(b) We need to use the power rule to find that <math>\int_2^5 4x - 5 \, dx = 2x^2 - 5x \Bigr|_2^5</math>
 
 
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|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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!(a) Step 2: &nbsp;
 
|-
 
|-
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|(a)
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|Now, we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
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::<math>\frac{1}{6} \int e^{u}\, du = \frac{1}{6}e^u.</math>
 
 
|-
 
|-
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|(b) We just need to evaluate at the endpoints to finish the problem:
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| to find&nbsp; <math style="vertical-align: -60%">\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>
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<math>\begin{array}{rcl}2x^2 - 5x \Bigr|_2^5 & = & 2(5^2) - 5(5) -(2(2)^2 - 5(2)\\
 
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& = & 50 - 25 -(8 - 10)\\
 
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& = & 25 +2\\
 
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& = & 27
 
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\end{array}</math>
 
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
+
!(a) Step 3: &nbsp;
 
|-
 
|-
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|(a) Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
+
|Since this integral is an indefinite integral, we have to remember to add a constant&thinsp; <math style="vertical-align: 0%">C</math> at the end.
−
|-
 
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| to find&nbsp; <math>\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>
 
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 4: &nbsp;
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!(b): &nbsp;
 +
|-
 +
|Unlike part (a), this requires no substitution.  We can integrate term-by-term to find
 +
|-
 +
|
 +
::<math>\int_2^5 4x - 5 \, dx = 2x^2 - 5x \Bigr|_{x\,=\,2}^5.</math>
 
|-
 
|-
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|Since this integral is an indefinite integral we have to remember to add a constant&thinsp; <math style="vertical-align: 0%">C</math> at the end.
+
|Then, we evaluate:
 +
|-
 +
|
 +
::<math>\begin{array}{rcl}2x^2 - 5x \Bigr|_{x\,=\,2}^5 & = & 2(5^2) - 5(5) -(2(2)^2 - 5(2))\\
 +
& = & 50 - 25 -(8 - 10)\\
 +
& = & 25 +2\\
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& = & 27.
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\end{array}</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
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|(a)
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|'''(a)'''
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::<math>\frac{e^{3x^2 + 1}}{6} + C.</math>
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::<math>\int xe^{3x^2+1}\,dx\,=\,\frac{e^{3x^2 + 1}}{6} + C.</math>
 +
|-
 +
|'''(b)'''
 
|-
 
|-
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|(b) <math>27</math>
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|
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::<math>\int_2^54x - 5\,dx\,=\,27.</math>
 
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|}
  
 
[[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:02, 17 May 2015

Find the antiderivatives:

(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int xe^{3x^2+1}\,dx.}


(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_2^54x - 5\,dx.}
Foundations:  
This problem requires Integration by substitution (u - sub): If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = g(x)}   is a differentiable functions whose range is in the domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} , then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int g'(x)f(g(x)) dx \,=\, \int f(u) du.}
We also need our power rule for integration:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,}   for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\neq 0} .

 Solution:

(a) Step 1:  
(a) Use a u-substitution with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 3x^2 + 1.} This means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du = 6x\,dx} , or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dx=du/6} . Substituting, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x e^{3x^2+1}\,dx \,=\, \int xe^{u}\cdot\frac{du}{6}\,=\,\frac{1}{6}\int e^u\,du\,=\,\frac{1}{6}u. }
(a) Step 2:  
Now, we need to substitute back into our original variables using our original substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 3x^2 + 1}
to find  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.}
(a) Step 3:  
Since this integral is an indefinite integral, we have to remember to add a constant  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} at the end.
(b):  
Unlike part (a), this requires no substitution. We can integrate term-by-term to find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_2^5 4x - 5 \, dx = 2x^2 - 5x \Bigr|_{x\,=\,2}^5.}
Then, we evaluate:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl}2x^2 - 5x \Bigr|_{x\,=\,2}^5 & = & 2(5^2) - 5(5) -(2(2)^2 - 5(2))\\ & = & 50 - 25 -(8 - 10)\\ & = & 25 +2\\ & = & 27. \end{array}}
Final Answer:  
(a)
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int xe^{3x^2+1}\,dx\,=\,\frac{e^{3x^2 + 1}}{6} + C.}
(b)
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_2^54x - 5\,dx\,=\,27.}

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